package com.wc.codeforces.思维.Bewitching_Stargazer;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

/**
 * @Author congge
 * @Date 2025/1/5 22:42
 * @description
 * https://codeforces.com/contest/2053/problem/C
 */
public class Main {
    /**
     * 思路：
     * 等差数列 a[n] + a[m] = a[p] + a[q], n + m == p + q
     * 可以观察每一段被分开的时候是 左右两边被选的位置都是中心位置, 并且相加正好是 a[1] + a[n] = n + 1
     * 所以计算有多少对, 可以利用位运算的思想
     */
    static FastReader sc = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);
    static int n, k;

    public static void main(String[] args) {
        int T = sc.nextInt();
        while (T-- > 0) {
            n = sc.nextInt();
            k = sc.nextInt();
            // res存储的是对数
            long mul = n + 1, res = 0, num = 1;

            while (n >= k) {
                if ((n & 1) == 1) res += num;
                n >>= 1;
                // 因为左右同时有所以 *= 2
                num <<= 1;
            }
            out.println(mul * res / 2);
        }
        out.flush();
    }
}

class FastReader {
    StringTokenizer st;
    BufferedReader br;

    FastReader() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }

    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    int nextInt() {
        return Integer.parseInt(next());
    }

    String nextLine() {
        String s = "";
        try {
            s = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return s;
    }

    long nextLong() {
        return Long.parseLong(next());
    }

    double nextDouble() {
        return Double.parseDouble(next());
    }

    // 是否由下一个
    boolean hasNext() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                String line = br.readLine();
                if (line == null)
                    return false;
                st = new StringTokenizer(line);
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return true;
    }
}

